Derivatives of power and constant functions Find the derivative of the following functions.

$y={x}^{5}$

IMLOG10ct
2021-11-26
Answered

Derivatives of power and constant functions Find the derivative of the following functions.

$y={x}^{5}$

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Forneadil

Answered 2021-11-27
Author has **18** answers

Find $\frac{d}{dx}\left({x}^{5}\right)$

Apply the power rule$\frac{d}{dx}\left({x}^{n}\right)=n{x}^{n-1}\text{with}n=5:$

$$\left(\begin{array}{c}\frac{d}{dx}({x}^{5})\end{array}\right)=(5{x}^{4})$$

Thus,$\frac{d}{dx}\left({x}^{5}\right)=5{x}^{4}$

Answer:

$\frac{d}{dx}\left({x}^{5}\right)=5{x}^{4}$

Apply the power rule

Thus,

Answer:

asked 2021-10-05

Find the derivatives of the following functions.

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$\underset{x\to \frac{\pi}{2}}{lim}(1-\frac{\pi}{2})\mathrm{tan}(x)$

I'm trying to putting $x=t+\pi /2$ zero is in denominator

I'm trying to putting $x=t+\pi /2$ zero is in denominator

asked 2022-06-05

The function,

$f(x)=\{\begin{array}{ll}({e}^{x}-1)/x& x\ne 0\\ 1& x=0\end{array}$

is continuous and differentiable at $x=0$. By composition, the $x\ne 0$ case is analytic everywhere it is defined, and integrating $f$ around 0 in the complex plane yields 0 and suggests that there is no pole there. Furthermore, working from the Taylor series of ${e}^{x}$, the $x\ne 0$ case can be seen to have a series that is defined at 0 and equal to 1:

$\frac{{e}^{x}-1}{x}=\frac{-1+\sum _{k=0}^{\mathrm{\infty}}{x}^{k}/k!}{x}=\frac{\sum _{k=1}^{\mathrm{\infty}}{x}^{k}/k!}{x}=\sum _{k=0}^{\mathrm{\infty}}\frac{1}{k+1}\frac{{x}^{k}}{k!}$

From this we can also see that $f$ is analytic and infinitely differentiable.

All taken together, it seems strange that such a nicely behaved function would have an irremovable "patch" necessary at $x=0$. Nonetheless, I have not been able to come up with another way to write the function that allows it to be defined at zero without a special case there. Is it impossible to write this analytic function without a case block?

$f(x)=\{\begin{array}{ll}({e}^{x}-1)/x& x\ne 0\\ 1& x=0\end{array}$

is continuous and differentiable at $x=0$. By composition, the $x\ne 0$ case is analytic everywhere it is defined, and integrating $f$ around 0 in the complex plane yields 0 and suggests that there is no pole there. Furthermore, working from the Taylor series of ${e}^{x}$, the $x\ne 0$ case can be seen to have a series that is defined at 0 and equal to 1:

$\frac{{e}^{x}-1}{x}=\frac{-1+\sum _{k=0}^{\mathrm{\infty}}{x}^{k}/k!}{x}=\frac{\sum _{k=1}^{\mathrm{\infty}}{x}^{k}/k!}{x}=\sum _{k=0}^{\mathrm{\infty}}\frac{1}{k+1}\frac{{x}^{k}}{k!}$

From this we can also see that $f$ is analytic and infinitely differentiable.

All taken together, it seems strange that such a nicely behaved function would have an irremovable "patch" necessary at $x=0$. Nonetheless, I have not been able to come up with another way to write the function that allows it to be defined at zero without a special case there. Is it impossible to write this analytic function without a case block?

asked 2021-10-22

Determine whether statement is true or false, and explain why.

The equation y= 3x+4 represents the equation of a line with slope 4.

The equation y= 3x+4 represents the equation of a line with slope 4.

asked 2022-05-23

Given is the sequence ${x}_{1}=0,\phantom{\rule{thickmathspace}{0ex}}{x}_{n+1}=\sqrt{2+{x}_{n}}$. Prove:

$\underset{n\to \mathrm{\infty}}{lim}{2}^{n}\sqrt{2-{x}_{n}}=\pi $

Hint:

Use the following formulas:

$\mathrm{cos}\left(\frac{x}{2}\right)=\sqrt{\frac{1+\mathrm{cos}x}{2}}$

$\mathrm{sin}\left(\frac{x}{2}\right)=\sqrt{\frac{1-\mathrm{cos}x}{2}}$

Any idea how to solve this problem?

$\underset{n\to \mathrm{\infty}}{lim}{2}^{n}\sqrt{2-{x}_{n}}=\pi $

Hint:

Use the following formulas:

$\mathrm{cos}\left(\frac{x}{2}\right)=\sqrt{\frac{1+\mathrm{cos}x}{2}}$

$\mathrm{sin}\left(\frac{x}{2}\right)=\sqrt{\frac{1-\mathrm{cos}x}{2}}$

Any idea how to solve this problem?

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Find the derivatives of the functions $y=\sqrt{2x}\mathrm{sin}\left(\sqrt{x}\right)$ .

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Solving the following limit without LHospitals rule: $\underset{x\to 0}{lim}\frac{\mathrm{sin}({x}^{2}+2)-\mathrm{sin}(x+2)}{x}$